Find the value of ∈ , 4(2 − 1) > 24
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Let f be continuous on a closed interval [a,b] and differentiable on the open interval (a,b). If f(a)=f(b), then there is at least one point c in (a,b) wheref
′
(c)=0.
Given f=x
3
−3x in [−
3
,0]
⇒f
′
(x)=3x
2
−3
f
′
(c)=3c
2
−3=0
⇒c
2
=1
⇒c=−1
as c is in [−
3
,0]
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